p0372 - z = 1 2 H and A = 3 H 2 , wherefore F x B = g zA =...

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3.72. CHAPTER 3, PROBLEM 72 329 3.72 Chapter 3, Problem 72 Figure 3.57: Equivalent problems for a two-layer fluid. To begin, use superposition as illustrated in Figure 3.57. For Problem A, the arc’s projection on a vertical plane is a rectangle of height H and width 3 H . So, the centroid is located at z = 3 2 H and A =3 H 2 .Thu s , F x A = ρ g zA = 9 2 ρ gH 3 and F z A = ρ gH 2 (3 H )+ 1 2 ρ gH 2 (3 H )= 9 2 ρ gH
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Unformatted text preview: z = 1 2 H and A = 3 H 2 , wherefore F x B = g zA = 3 2 gH 3 and F z B = g w 1 2 H 2 W (3 H ) = 3 2 gH 3 So, the solution to the original problem is F x = F x A + F x B = 9 2 gH 3 + 3 2 gH 3 = 6 gH 3 F z = F z A + F z B = 9 2 gH 3 + 3 2 gH 3 = 6 gH 3 Therefore, the force vector is F = 6 gH 3 ( i + k )...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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